Structural Engineering & Statics
where the load goes
Concept

Redundancy — where it appears

How many more restraints or members a structure has than equilibrium needs, which is what makes its forces depend on stiffness. It buys an alternative load path and costs a stiffness calculation, and it makes the structure sensitive to settlements and temperature that statics cannot see.

Named by 12 essays across 4 fields — each of them below, with the objects they name alongside it.

Counting unknowns against equations. Three frames differing by one member. Two equilibrium equations per joint, one unknown per member and one per restraint: fewer unknowns than equations is a mechanism, equal is solvable by statics, more needs stiffness.

Counting the unknowns, and finding out whether statics can answer

Two equations per joint, one unknown per member, one per restraint. Subtract, and the sign of the answer says whether the structure is a mechanism, solvable, or beyond what equilibrium alone can settle.

equilibrium · Determinacy Take that one away and the load finds another route. A 6-panel pratt truss under 20 kN at each top node, before and after member 2 is removed. The load redistributes. The worst-affected survivor now carries 2.03 times what it did, and four members that carried nothing before are now working. Whether that is survival depends on how much spare capacity was there, which is a different question from whether the frame was strong enough.

The structure that survives losing a member

Every check in this collection asks what a structure carries. None of them asks what is left when part of it is gone — and two frames with the same members, the same weight and the same factor of safety can answer that question completely differently.

structures · Robustness A frame with no plane to be drawn in. The tetrahedron, solved: three equations at every free joint, one axial force in every member. Members drawn heavy carry more; tension and compression are separated by the sign that came back from the solve rather than by inspection. The count is m + r = 12 against 3j = 12, which makes it exactly determinate, and joint equilibrium closes to 1.8e-15.

Three equations at every joint

A plane truss is determinate when m + r = 2j. A space frame needs 3j, and that one changed digit is why a cube of twelve bars is six mechanisms short while looking perfectly solid — and why every three-dimensional frame ever built is made of triangles in several planes at once.

structures · Space frame The tie is a redundancy, so its stiffness decides the thrust. Thrust and rib bending for a 60 m tied arch of 0.15 rise ratio, against the stiffness of its tie. Cut the tie and the structure is a curved simply supported beam, so the tie force is the one redundant and the force method gives it: with a rigid tie the answer is 2229 kN, within 0.9 per cent of the funicular wL²/8f, and the shortfall is the arch's own axial shortening. A real tie stretches 68 mm and returns 2168 kN — 2.7 per cent of the flexibility is the tie — and whatever thrust the arch does not get, it carries as bending: 739 kNm at 30 m. A tenth of the tie stiffness is not a tenth of the problem; it is a different structure.

The thrust that never reaches the ground

Every arch on this site has ended at the same sentence — the foundation is where an arch is really decided. A tie changes the sentence without changing the arithmetic: the horizontal force is still there, still the same size, and it now closes on itself through a bar at deck level.

structures · Tied arch The two theorems close on the answer from opposite sides. A pinned-base portal frame under 200 kN at mid-span and 80 kN at the eaves, with a plastic moment of 200 kNm. The three horizontal lines are mechanisms — work equations, each an upper bound: the beam mechanism gives 1.000, sway 1.250 and the combined one 0.714. The curve is the lower bound: for each value of the one redundant, the largest load factor whose moment field stays inside the plastic moment everywhere. Its peak is 0.714, and it touches the lowest mechanism exactly — so the collapse load is known rather than bracketed. Using the beam mechanism instead would have claimed 1.40 times the real capacity, and the sway one 1.75 times, both of them on the wrong side.

Two ways of being wrong

Plastic analysis has two theorems and they point in opposite directions. Any equilibrium field that nowhere exceeds the plastic moment gives a load at or below the collapse load; any mechanism gives one at or above it. Only one of those errors is safe.

equilibrium · Bound theorems The check that everything adds up, and the error it cannot see. Four versions of the same 3-bay, 4-storey frame, with the global equilibrium residual each one produces — the sum of the reactions against the sum of the applied loads, as a fraction of the applied total. It is the first thing every analysis prints and it is worth having: a lost restraint and a load entered in the wrong unit both show up immediately, at 8% and 32%, because both change what the structure is carrying. The fourth bar is the point. A member whose stiffness is wrong by a factor of ten redistributes the internal forces completely — the second bar shows the change in the member forces, 24% — and the global residual is exactly zero, because the wrong answer is still in equilibrium with the same loads. Equilibrium is one equation per degree of freedom of the whole body, and a stiffness error lives entirely in the many equations underneath it. A model can satisfy every equilibrium check ever devised and be a model of a different structure.

The check that cannot see the error

Every analysis prints a global equilibrium residual, and it is the first thing anybody looks at. It catches a lost restraint and a load entered in the wrong unit immediately. It is structurally incapable of catching a member whose stiffness is wrong by a factor of ten, because the wrong answer is still in equilibrium with the same loads.

equilibrium · Equilibrium check The answer arrives in instalments. The hogging moment at support 1 of a three-span beam, cycle by cycle. It starts at the fixed-end moment of 98.0 kNm — the value with every joint clamped — and settles at 156.9 kNm against an exact 156.9. The error falls by about a factor of four per cycle: 21.03, 5.92, 1.54, 0.60 kNm after one, two, three and four. Two cycles is an engineering answer and nobody had to invert anything.

Why it converges, and how fast

Moment distribution is an iteration, and iterations do not always converge. This one always does, at a rate the beam's own proportions fix — about a factor of four per cycle on a regular beam and considerably worse on an irregular one, which is where the method's reputation for two cycles being enough comes from and where it stops being true.

deflection · Moment distribution Removing each member in turn. Every member of a 8-panel pratt truss removed one at a time, with the worst demand on the survivors plotted against the member removed. Four of the 35 leave a mechanism — the bars drawn to the top of the frame — and for those there is no redistribution to compute, because there is no structure left. The rest redistribute, and the worst of them asks a survivor for 2.04 times what it carried before. A single number for robustness does not exist: it depends on which member goes.

A determinate truss has no robustness at all

Remove any one of a Warren truss's thirty-one members and what is left is a mechanism. Not weakened — gone, with no set of forces that holds the load in any position. Robustness is not a property a structure has by degree; it is bought by adding members that carry nothing until something else stops carrying, and a truss without them has none of it to measure.

structures · Robustness What each missing bolt costs. The weakest-direction capacity of six bolts at 75 by 75 mm, loaded through a point (150, 150) mm from the centroid, with the whole group and with each bolt in turn left out. The full group carries 135.8 kN. Leaving out bolt 3 leaves 90.2 kN, a loss of 34 per cent; leaving out bolt 2 leaves 131.1 kN, a loss of 4. One sixth of the bolts is not one sixth of the capacity, and which sixth it is matters by a factor of 10. The dashed line is the capacity a group that lost a proportional share would have, 113.2 kN.

The bolt that was never fitted

A six-bolt bracket found with five bolts in it has lost a sixth of its fasteners and between four and thirty-four per cent of its capacity, depending which one is missing. The share is the smallest of the three things that changed: the centroid moves away from the gap, which lengthens the load's own lever arm, and the polar moment falls by more than the count does. The bolt whose absence costs most is not the bolt that governed the check.

connections · Bolt group Four ways to put six bolts in one plate. Six bolts inside a 150 × 150 mm field of bolt centres, no two closer than 60 mm, loaded through a point (200, 0) mm from the field's centre. The two-column layout carries 193.5 kN complete and 138.8 kN with its worst bolt missing. The ring carries 167.2 kN complete and 117.8 kN with its worst bolt missing. The strongest found carries 234.3 kN complete and 149.7 kN with its worst bolt missing. The most robust found carries 231.6 kN complete and 174.7 kN with its worst bolt missing. The layout found by maximising the complete capacity and the layout found by maximising the worst omission are different layouts, 1 per cent apart when complete and 17 per cent apart with a bolt missing, and both beat the ring — the most evenly spread of the four — on both counts. The dashed line runs from each group's centroid to the load: 200 mm, 200 mm, 180 mm, 188 mm. The ringed bolt is the one each group can least afford to lose.

The strongest layout leans on one bolt

Search a plate for the six bolt positions that carry most and the answer carries 234 kN — and loses 36 per cent of it if one particular bolt is missing. Move that one bolt fifty millimetres, into the corner the optimum had just left, and the group carries 232 kN and loses 25 per cent whichever bolt goes. Robustness here costs one per cent of strength, and a search for strength alone will never find it.

connections · Bolt group Fully stressed, with a choice of diagonals. Two trusses sized so that every member is at the allowable stress under the full load, each member drawn as wide as its area, tension and compression in two colours. Above, an eight-panel Pratt truss as deep as a panel is long, with a counter-diagonal in each of its six interior panels, loaded by 10 at every bottom joint, found by resizing and reanalysing until nothing changes; below, the same truss without its counters. The counters in the two middle panels have shrunk to nothing (dotted) and the four nearer the supports have stayed, working in compression beside the diagonals in tension. The truss that kept them needs 1,200 units of steel against the Pratt's 1,220, and deflects 24.0 at mid-span against 26.0.

The truss whose forces follow its sections

In a determinate truss each member's force is fixed before its section is chosen, so sizing for strength and sizing for stiffness can be done in either order. Put a counter-diagonal in every panel and they cannot. The fully stressed design becomes an iteration that starves some counters to nothing and keeps others, lands on a different truss from every start, and — whichever it lands on — weighs the same and deflects the same. Then enlarge one group of members to stiffen it, and a vertical nobody touched is overloaded by 58 per cent.

deflection · Truss deflection Three members after a diagonal goes. The forces in three members of the counter-braced truss when the diagonal 9–2, carrying 242 kN, is removed instantaneously, with 2 per cent damping, over one and a half of the damaged truss's first periods (0.57 s); dashed, the static force each settles to. The counter-diagonal 1–10 goes from −112 kN to −354 kN and peaks at −491 kN, 1.57 times its change. The bottom chord 3–4 ends exactly where it started, 750 kN, and on the way peaks at 1,115 kN. The bottom chord 2–3 settles lower, at 556 kN, after a first swing the other way, to 905 kN.

The factor of two belongs to one mode

A member that fails suddenly hands its force to the structure around it all at once, and the convention is to double the static answer: a load applied suddenly to a spring overshoots to twice its static deflection. A truss is not one spring. Take a diagonal out of a counter-braced truss in an instant and some members swing to three times their change of force, one swings the wrong way first, and a bottom chord whose force does not change at all passes through half as much again as it carries — because every mode overshoots by two, at its own time, and a member is a sum of modes.

structures · Robustness

Named alongside it

The objects these essays reach for when they reach for this one.

RobustnessDeterminacyMechanismCompatibilityDuctilityLoad pathTrussAlternative load pathBolt groupEccentricityElastic methodEquilibrium

All concepts